# Scheme theoretically, when the union of the interserction is the intersection of the union

We have the definition:

Definition. Let $$X$$ be a scheme. Let $$Z,Y⊂X$$ be closed subschemes corresponding to quasi-coherent ideal sheaves $$\mathcal{I},\mathcal{J}⊂\mathcal{O}_X$$. The scheme theoretic intersection of $$Z$$ and $$Y$$ is the closed subscheme of $$X$$ cut out by $$\mathcal{I}+\mathcal{J}$$. The scheme theoretic union of Z and Y is the closed subscheme of $$X$$ cut out by $$\mathcal{I}∩\mathcal{J}$$.

Is true, in general, that given closed subschemes $$Y,Z$$ and $$W$$ of $$X$$ we have

$$Y\cap (Z\cup W)= (Y\cap Z)\cup (Y\cap W), \mbox{scheme-theoretically?}$$

If not, there is any necessary and sufficient conditions? Any hint, reference or solution is welcome!!

Remark: Using the language of ideal, if we denote $$I_*$$ the ideal of the variety $$*$$ we have just one inclusion, in general $$I_Y+(I_Z\cap I_W)\subseteq (I_Y+I_Z)\cap(I_Y+I_W).$$

• As you say, the corresponding statement about ideals is not true in general. A counterexample there translates to a counterexample in affine schemes. Aug 4, 2020 at 22:25
• Exactly @ZhenLin, but any hint about a necessary and sufficient condition for this be true?
– IMP
Aug 4, 2020 at 22:55
• For example, again, talking about ideals. If the deals $I_Y+I_Z$ and $I_Y+I_W$ are co-prime, that is, if they sum up to the unitary ideal. Then the statement is true. In other words it is a sufficient condition.
– IMP
Aug 4, 2020 at 23:18

taking $$V(y^2-x^2),V(y)\subset \Bbb{A}^2$$ we see that $$V(x^2-y^2) = V(x-y)\cup V(x+y)$$. If we take $$X= V(x-y), Y = V(x+y), Z = V(y)$$ we see that $$(X\cup Y)\cap Z =\text{Spec }k[x,y]/(x^2-y^2,y) = \text{Spec }k[x,y]/(x^2 ,y) \cong \text{Spec } k[x]/(x^2) \tag{1}$$
While $$(X\cap Z)\cup (Y\cap Z) = \text{Spec } k[x,y]/(x,y) \cong \text{Spec } k \tag{2}$$